- #1
tpingt
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I'm relatively new to calculus, and this question was bugging me, so I have decided to ask it.
We have the function [tex]y=1/x[/tex] with domain [tex]x\geq1[/tex] and we rotate the curve around the x-axis in order to form a solid of revolution. (Gabriel's Horn)
The integral is [tex]V=\pi\int^{z}_{1}1/x^2 dx[/tex] and we evaluate to [tex]V=\pi(1-1/z)[/tex]
Take the limit as z approaches infinity: [tex]\lim_{z \to \infty}\pi(1-1/z)=\pi[/tex]
Apparently my teacher says that the volume is a finite amount, which is to say [tex]\pi[/tex]. However isn't pi an irrational number? Meaning that its digits keep going without end? If pi is not finite, then how can the volume be finite? Wouldn't the volume keep getting minutely closer to pi as x tends to infinity?
Thanks!
We have the function [tex]y=1/x[/tex] with domain [tex]x\geq1[/tex] and we rotate the curve around the x-axis in order to form a solid of revolution. (Gabriel's Horn)
The integral is [tex]V=\pi\int^{z}_{1}1/x^2 dx[/tex] and we evaluate to [tex]V=\pi(1-1/z)[/tex]
Take the limit as z approaches infinity: [tex]\lim_{z \to \infty}\pi(1-1/z)=\pi[/tex]
Apparently my teacher says that the volume is a finite amount, which is to say [tex]\pi[/tex]. However isn't pi an irrational number? Meaning that its digits keep going without end? If pi is not finite, then how can the volume be finite? Wouldn't the volume keep getting minutely closer to pi as x tends to infinity?
Thanks!